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2 years ago

Are -3.86 and 2.9 on the same side or the opposite side of zero ?

Mathematics

1 answer:

RoseWind [281]2 years ago

8 0

-3.89 and 2.9 are on opposite sides

-3.89 is a negative number, so it is to the left of 0

2.9 is a positive number, so it is to the right of 0

Hope this helps

-AaronWiseIsBae

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2 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

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3 years ago

Describe the transformation in the figure below

Keith_Richards [23]

Answer:

A) Reflection.

Step-by-step explanation:

From the figure shown, we can see that the blue triangle is flipped about the x-axis to form the red triangle.

When an object is flipped about a line, then this transformation is called reflection about that line.

Thus, we can say that reflection is taking place in the figure shown in which the x-axis is acting like a mirror. The blue triangle is the pre-image and the red triangle is the image after reflection.

The reflection about x-axis is given by:

$Pre-image(x,y)\rightarrow Image(x,-y)$

We can check the co-ordinates of the triangles and confirm the the transformation occurring

Points of pre-image triangle:

$(0,2),(4,5),(6,2)$

Points of image triangle

$(0,-2),(4,-5),(6,-2)$

Thus, the transformation is reflection.

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3 years ago